Math problem based on real life

[MotionMan]

I was having a discussion with another basketball coach about a situation in our league and the math of our issue left us both stumped:

-The league has 6 teams of 10 players each (60 players total);
-3 of the players are bound to Team A (kids of Team A's coaches);
-10 parents of other players requested to be on Team A (which is allowed but not guaranteed);
-None of those 10 players are assigned to Team A.

Assuming a purely random draw, what are the odds that none (0) of the 10 players (out of the 57 remaining players) would end up on Team A (filling the remaining 7 spots on the team)?

Thanks.

MotionMan

 

[MagnusTheBrewer]

My math may be rusty but, if the 10 slots requested have the same chance as every one else, it would be 7 chances in 57 or 12.2% so, the odds of NOT getting a slot are ~87.8%

 

[deadlyapp]

My math may be rusty but, if the 10 slots requested have the same chance as every one else, it would be 7 chances in 57 or 12.2% so, the odds of NOT getting a slot are ~87.8%

Although the odds technically change each time a player is chosen. Depending on what order the 10 are placed, the odds can get either better or worse. I don't technically think that is probability though.

 

[MotionMan]

My math may be rusty but, if the 10 slots requested have the same chance as every one else, it would be 7 chances in 57 or 12.2% so, the odds of NOT getting a slot are ~87.8%

That seems like 1/2 of the equation (the odds of any player not getting on Team A). We thought the number of requesters (10) would factor in, too.

MotionMan

 

[MagnusTheBrewer]

Although the odds technically change each time a player is chosen. Depending on what order the 10 are placed, the odds can get either better or worse. I don't technically think that is probability though.

I should know by now, brandy and math don't mix. You're right.

 

[MotionMan]

Although the odds technically change each time a player is chosen. Depending on what order the 10 are placed, the odds can get either better or worse. I don't technically think that is probability though.

Right, except I am mostly concerned with any one of the 10, not all of them.

The two choices are 1) zero requesters on Team A and 2) at least one requester on Team A.

MotionMan

 

[MagnusTheBrewer]

That seems like 1/2 of the equation (the odds of any player not getting on Team A). We thought the number of requesters (10) would factor in, too.

MotionMan

Only if they're given some preference. Otherwise, they're exactly the same as any other possibility.

 

[MotionMan]

Only if they're given some preference. Otherwise, they're exactly the same as any other possibility.

Except we are looking for the probability of none of the requesters being chosen. It seems like 10 vs. 47, no?

MotionMan

 

[Fayd]

I was having a discussion with another basketball coach about a situation in our league and the math of our issue left us both stumped:

-The league has 6 teams of 10 players each (60 players total);
-3 of the players are bound to Team A (kids of Team A's coaches);
-10 parents of other players requested to be on Team A (which is allowed but not guaranteed);
-None of those 10 players are assigned to Team A.

Assuming a purely random draw, what are the odds that none (0) of the 10 players (out of the 57 remaining players) would end up on Team A (filling the remaining 7 spots on the team)?

Thanks.

MotionMan

https://www.wolframalpha.com/input/?i=((10choose0)(47choose7))/(57choose7)

~24%.

simple hypergeometric probability problem.

 

[Modular]

https://www.wolframalpha.com/input/?i=((10choose0)(47choose7))/(57choose7)

~24%.

simple hypergeometric probability problem.

Such a badass, underused website.

 

[Fayd]

Such a badass, underused website.

lol, true, but you don't need the website for this. i just didn't have the calculator handy to do nCr's

 

[Matthiasa]

https://www.wolframalpha.com/input/?i=((10choose0)(47choose7))/(57choose7)

~24%.

simple hypergeometric probability problem.

Yes because that is how teams are picked... one team picks all their players then the next team does... and so on.

 

[Fayd]

Yes because that is how teams are picked... one team picks all there players then the next team does... and so on.

he asked for a purely random draw for a given team.

given that, my math is sound.

we could reduce the second column if we knew how many kids were tied to other teams, but that information was not contained in the question.

 

[her209]

(47/57) x (46/56) x (45/55) x ... x (41/51) = (47x46x45x44x43x42x41) / (57x56x55x54x53x52x51) = 23.7%

 

[MotionMan]

he asked for a purely random draw for a given team.

given that, my math is sound.

we could reduce the second column if we knew how many kids were tied to other teams, but that information was not contained in the question.

That is a good point. I forgot about that. There are 5 other teams and each of them probably have 2 tied players.

BTW, I am not sure if the teams are set one team at a time or on a round-robin basis.

MotionMan

 

[her209]

That is a good point. I forgot about that. There are 5 other teams and each of them probably have 2 tied players.

BTW, I am not sure if the teams are set one team at a time or on a round-robin basis.

MotionMan

If each team as at least 2 tied players, with Team A having 3, then the math would be

(37x36x35x34x33x32x31) / (47x46x45x44x43x42x41) = 16.37%

 

[MotionMan]

If each team as at least 2 tied players, with Team A having 3, then the math would be

(37x36x35x34x33x32x31) / (47x46x45x44x43x42x41) = 16.37%

That seems more like it.

So, the chances of Team A not getting even one of the 10 requesters is less than 17%. Seems very unlikely, no?

MotionMan

 

[Fayd]

That seems more like it.

So, the chances of Team A not getting even one of the 10 requesters is less than 17%. Seems very unlikely, no?

MotionMan

no.

that's not statistically significant for any measure. you have a 1/6 chance of getting the draw you did.

anyways, yeah, hypergeom gets same answer as her209

https://www.wolframalpha.com/input/?i=((10choose0)(37choose7))/(47choose7)

it just seems easier than his method.

 

[Red Squirrel]

My solution to problems like these is to write a C++ program that randomizes and keeps stats on results and can be run in a user defined amount of times, or for something that has finite possibilities, each single possibility is tried. While the math way is more efficient, the C++ way is easier.

 

[twinrider1]

Your solution is not real life.

This is real life....

Separate pool for the 10 kids that want to be on Team A. Randomly draw 7 names. The remaining 3 go back to the main pool.

Now draft, or random draw, or whatever the standard practice is, for the remaining 50 kids to the remaining 9 teams.

 

[her209]

My solution to problems like these is to write a C++ program that randomizes and keeps stats on results and can be run in a user defined amount of times, or for something that has finite possibilities, each single possibility is tried. While the math way is more efficient, the C++ way is easier.

Orrrrrrrrr you could just write a C++ program that enumerates all possible outcomes and determines the number of times when none of the 10 kids are on Team A divided by the number of possible outcomes.

 

[Paperdoc]

OP did not provide an important part of the info. When making assignments of players to teams, do we pick ALL of Team A's people before picking ones for Team B? Or, do we pick the first player for each of six teams all around, then go on to the second players, etc.?

Assuming the "pick all of Team A first" option, her209 got the right answer. The method is to ask what are the odds of NOT picking one of the "Special 10" each time?

There will be seven picks to make (3 already on Team A). For the first of these, the odds of NOT picking one of the "Special 10" are (57-10)/57. For the second pick, the odds are (56-10)/56. So the odds of NOT having any "Special 10" players picked after the second round is the product of those two fractions. Continue on for 7 picks and you get fractional odds of 0.237878, or 23.8% odds of having NONE of the "Special 10" on Team A.

If assignments are done the other way, it gets trickier to work out, and I have not done that.

 

[MotionMan]

OP did not provide an important part of the info. When making assignments of players to teams, do we pick ALL of Team A's people before picking ones for Team B? Or, do we pick the first player for each of six teams all around, then go on to the second players, etc.?

Assuming the "pick all of Team A first" option, her209 got the right answer. The method is to ask what are the odds of NOT picking one of the "Special 10" each time?

There will be seven picks to make (3 already on Team A). For the first of these, the odds of NOT picking one of the "Special 10" are (57-10)/57. For the second pick, the odds are (56-10)/56. So the odds of NOT having any "Special 10" players picked after the second round is the product of those two fractions. Continue on for 7 picks and you get fractional odds of 0.237878, or 23.8% odds of having NONE of the "Special 10" on Team A.

If assignments are done the other way, it gets trickier to work out, and I have not done that.

BTW, I am not sure if the teams are set one team at a time or on a round-robin basis.

The teams are selected by the league staff.

I have left out two of the complicating factors, which are that the teams are suppose to be somewhat even on skill and evenly distributed based on height. Though the 10 requestors are fairly evenly distributed on those factors, they should not be used as part of the equation.

MotionMan

 

[DrPizza]

OP did not provide an important part of the info. When making assignments of players to teams, do we pick ALL of Team A's people before picking ones for Team B? Or, do we pick the first player for each of six teams all around, then go on to the second players, etc.?

Assuming the "pick all of Team A first" option, her209 got the right answer. The method is to ask what are the odds of NOT picking one of the "Special 10" each time?

There will be seven picks to make (3 already on Team A). For the first of these, the odds of NOT picking one of the "Special 10" are (57-10)/57. For the second pick, the odds are (56-10)/56. So the odds of NOT having any "Special 10" players picked after the second round is the product of those two fractions. Continue on for 7 picks and you get fractional odds of 0.237878, or 23.8% odds of having NONE of the "Special 10" on Team A.

If assignments are done the other way, it gets trickier to work out, and I have not done that.

You're overthinking it. It's the same.

Think of it this way: the teams take turns picking. After all the players have been picked, you line the players along the wall, one team at a time. If team A was players 31 through 40, you'd be asking, what's the probability that 31 through 40 had none of the players from the group that wanted to be on that team.

 

[DCal430]

The odds that no one will go to team A is 0.1957 most, odds and probability are not the same.